In This post you will get answer of Q.
A conducting loop (as shown) has total resistance R. A uniform magnetic field $B = gamma$ t is applied perpendicular to plane of the loop where $gamma$ is a constant and t is time. The induced current flowing through loop is
Solution:
Given,
uniform magnetic field $(B)=gamma ,t$

Total flux, $phi = B _{1} ,A _{1}+ B _{2} ,A _{2} $
$=B b^{2} ,cos 0^{circ}+B a^{2} ,cos, 180^{circ}$
$=B b^{2}-B a^{2} $
$ phi =Bleft(b^{2}-a^{2}right)$
$ phi =gamma tleft(b^{2}-a^{2}right),,,,,dots(i)$
We know that,
induced current (i) $=frac{|e|}{R}=frac{left|frac{d phi}{d t}right|}{R}$
From Eq. (i),
$i=frac{frac{d}{d t}left[gamma tleft(b^{2}-a^{2}right)right]}{R}$
$i=frac{left(b^{2}-a^{2}right) gamma}{R}$
Solution:
Given,
uniform magnetic field $(B)=gamma ,t$
Total flux, $phi = B _{1} ,A _{1}+ B _{2} ,A _{2} $
$=B b^{2} ,cos 0^{circ}+B a^{2} ,cos, 180^{circ}$
$=B b^{2}-B a^{2} $
$ phi =Bleft(b^{2}-a^{2}right)$
$ phi =gamma tleft(b^{2}-a^{2}right),,,,,dots(i)$
We know that,
induced current (i) $=frac{|e|}{R}=frac{left|frac{d phi}{d t}right|}{R}$
From Eq. (i),
$i=frac{frac{d}{d t}left[gamma tleft(b^{2}-a^{2}right)right]}{R}$
$i=frac{left(b^{2}-a^{2}right) gamma}{R}$